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        <p>感觉是很久没打Codeforces了，有点手生，前期卡题很严重。 C题明明有简单的贪心，却想了很久的DP，还是个错的，D题还读错了题，没读到取最大值，白想了很久。最后的G题感觉也有点超出能力范围了。 直接一手 <span class="math inline">\(-5\)</span> rating。永远不变的真理：不打等于上分。</p>
<a id="more"></a>
<h1 id="a-bovine-dilemma">A Bovine Dilemma</h1>
<h2 id="题意">题意</h2>
<p>给出 <span class="math inline">\(n\ (n\le50)\)</span> 个 <span class="math inline">\(x_i\)</span> 表示直角坐标系上，X正半轴上的 <span class="math inline">\(n\)</span> 个点 <span class="math inline">\((x_i,0)\)</span>。求从中任选两个点和 <span class="math inline">\((0,1)\)</span> 点能组多少个面积不同的三角形。</p>
<h2 id="题解">题解</h2>
<p>直接认为X正半轴上的点是三角形的底，那么三角形的高已经固定是 <span class="math inline">\(1\)</span> 了，此时只要判断有多少种不同的底即可，那么 <span class="math inline">\(n^2\)</span> 枚举即可。</p>
<h1 id="b-last-minute-enhancements">B Last minute enhancements</h1>
<h2 id="题意-1">题意</h2>
<p>给出 <span class="math inline">\(n\ (n\le10^5)\)</span> 个 <span class="math inline">\(a_i\ (a_i\le2\times10^5)\)</span>，每个数字可以保持原样或者 <span class="math inline">\(+1\)</span>，请问最多能构造出多少个不同的数字。</p>
<h2 id="题解-1">题解</h2>
<p>从小到大排序，对于当前的数字，如果在之前出现了，那么就 <span class="math inline">\(+1\)</span>，否则保持不变，这样贪心即可。</p>
<h1 id="c-canine-poetry">C Canine poetry</h1>
<h2 id="题意-2">题意</h2>
<p>给出一个长度不超过 <span class="math inline">\(10^5\)</span> 的字符串，可以任意更改每一位上的字符，求最少更改多少个字符，使得字符串中不包含长度大于 <span class="math inline">\(1\)</span> 的回文子串。</p>
<h2 id="题解-2">题解</h2>
<p>其实要求长度不大于 <span class="math inline">\(1\)</span> 是非常苛刻的，这意味着：</p>
<ul>
<li>相邻的两个字符不能相同，否则形如 <span class="math inline">\(aa\)</span> 的字串就是回文的。</li>
<li>隔一位相邻的两个字符不能相同，否则形如 <span class="math inline">\(aba\)</span> 的字串就是回文的。</li>
</ul>
<p>满足这两个条件即可，这样奇回文和偶回文都不存在了。</p>
<p>其实从左到右贪心地修改，每次使得当前字符和前两个字符不同即可，不需要考虑对后面的影响，因为小写字母有26个，总是能找到一个没有后效性的字符。 可以用一个 <span class="math inline">\(@\)</span> 字符表示绝对不会相同的字符。</p>
<h1 id="d-13th-labour-of-heracles">D 13th Labour of Heracles</h1>
<h2 id="题意-3">题意</h2>
<p>给出一颗规模为 <span class="math inline">\(n\ (n\le10^5)\)</span> 的树，树上的每个结点都有一个权值 <span class="math inline">\(w_i\ (w_i\le10^9)\)</span>。 现在可以对树的<strong>边</strong>进行 <span class="math inline">\(k\)</span> 染色，染色后：对于每种颜色，提取出该颜色的子图（该颜色的所有边以及这些边连接的点），子图中的每个连通块中所有点的权值和记为这个连通块的得分，所有连通块中最大得分是这个颜色的得分。 对于一种染色方案，所有颜色的得分求和就是这个染色方案的得分。 现在要求 <span class="math inline">\(k\)</span> 染色的最大得分，<span class="math inline">\(k\)</span> 从 <span class="math inline">\(1\)</span> 到 <span class="math inline">\(n\)</span> 都要输出。</p>
<h2 id="题解-3">题解</h2>
<p><span class="math inline">\(k\)</span> 从 <span class="math inline">\(1\)</span> 到 <span class="math inline">\(n\)</span> 都要输出，其实已经说明了这个题目需要动态地解决。 首先 <span class="math inline">\(1\)</span> 染色只有一种方案。当我们需要 <span class="math inline">\(2\)</span> 染色时，我们发现，如果第二种颜色和第一种颜色正好分开这棵树，两个子图都只有一个连通块时，染色方案的得分正好比 <span class="math inline">\(1\)</span> 染色多出那个两种颜色分割点的权值。而且不难发现，如果任意一种颜色的子图有多个连通块时，整体的得分将大打折扣。 所以，我们希望这个分割点的权值尽可能大，那就让这个点是整棵树权值最大的点就行了。 但是，如果这个点是个叶子结点，显然是不行的，所以，一个点能够充当多少次分割点，就是他的 <strong>度数</strong> <span class="math inline">\(-1\)</span>。 所以，给所有的点按权值从大到小排序，不断更新得分即可。</p>
<h1 id="e-apollo-versus-pan">E Apollo versus Pan</h1>
<p>## 题意</p>
<p>给出 <span class="math inline">\(n\ (n\le5\times10^5)\)</span> 个 <span class="math inline">\(x_i\ (x_i\le2^{60})\)</span>，求： <span class="math display">\[
\sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n (x_i\, \&amp; \, x_j) \cdot (x_j \, | \, x_k)
\]</span></p>
<h2 id="题解-4">题解</h2>
<p>这题看着吓人，<span class="math inline">\(5\times10^5\)</span>的规模下，要算一个 <span class="math inline">\(n^3\)</span> 的东西，其实不难。 当我们固定一个 <span class="math inline">\(a_j\)</span> 和 <span class="math inline">\(a_i\)</span> 时，我们发现后面的 <span class="math inline">\((x_j \, | \, x_k)\)</span> 其实是可以预处理出来的，它就只是个一个数字和所有数字的 <span class="math inline">\(and\)</span> 和而已，此时它就可以作为一个常数了，当我们只固定 <span class="math inline">\(a_j\)</span> 时发现，<span class="math inline">\((x_i \, \&amp; \, x_j)\)</span> 也是能预处理的，也能提取出来，那么整个式子就可以化为： <span class="math display">\[
\sum_{j=1}^n SumofAND_j \cdot SumofOR_j
\]</span> 那么这两个预处理可以按位拆分分别算贡献即可，整体复杂度是 <span class="math inline">\(O(nlogn)\)</span> 的。</p>
<h1 id="f-euclids-nightmare">F Euclid's nightmare</h1>
<h2 id="题意-4">题意</h2>
<p>给出 <span class="math inline">\(n\)</span> 个 <span class="math inline">\(m\)</span> 维的向量，这些向量的每一维都是只有 <span class="math inline">\(0,1\)</span> 的，向量做加法时，每一维求和都会 <span class="math inline">\(mod\ 2\)</span>。 特别地，每个向量最多有两维是 <span class="math inline">\(1\)</span> ，其余维度都是 <span class="math inline">\(0\)</span> 。 任取这 <span class="math inline">\(n\)</span> 个向量集合的一个子集求和可以得到一个向量，这些向量构成一个新的集合 <span class="math inline">\(T\)</span>，求 <span class="math inline">\(T\)</span> 集合的大小 <span class="math inline">\(|T|\)</span>。（特别的，如果子集为空，那么求出的向量为 <span class="math inline">\(0\)</span> 向量）。 同时，求出最小的子集 <span class="math inline">\(S&#39;\)</span>，使得这个子集能求出的新集合大小即为 <span class="math inline">\(|T|\)</span>，多个方案求字典序最小的方案。</p>
<h2 id="题解-5">题解</h2>
<p>首先我们可以知道，向量集合可以写成一个矩阵，矩阵中被化简掉的行是无效的，因为这些行可以被别的行表示。 突破口在于每个向量只有两维是 <span class="math inline">\(1\)</span>。 对于每个维度，它有三种状态，第一种是它还不能被表示出来，第二种是它能被表示出来，但它是和另一个维度捆绑的，第三种是它能被自由地表示出来。 那么我们没加入一个向量，就看他是否能更新我现有的这些状态，能就加入集合，不能就Skip。 整体的更新可以用一个并查集表示，自由的维度就和第 <span class="math inline">\(m+1\)</span> 个虚点连接即可。</p>

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